CHAPTER_IX.

GENERAL PROPERTIES OF EQUATIONS.

(238.) Any number or quantity which, when substituted for the unknown quantity in an equation, satisfies that equation, is called a root of that equation.

If the general algebraic equation

2-1

2

+An-1x+An=0

(1)

is satisfied by making x=a1, then a1 is a root of equation (1).

Substituting a for x in (1), we get

a1

is divisible by x-a1, consequently the left-hand member of (3), is divisible by x-u1.

Equation (3) does not differ from (1), since (3) was derived from (1) by subtracting from it equation (2), which is equal to 0. Therefore, equation (1) is also divisible by x-a1, hence the following property:

(239.) If a is a root of the general algebraic equation +An-1x+An=0,

then its left-hand member will be divisible by x-a1

As an example, suppose 3 is a root of the equation

23-7x2+36=0.

Now by the above property this equation must be divisible

be x-3.

Actually performing the division, we have

x3-7x2+36x-3 divisor.

by x-a1, we shall obtain for a quotient, a new equation of one degree less than equation (1), which may be represented as follows:

This equation must also have a root, which we will represent by a 2. Again, dividing (2) by x-a2, we shall obtain

a new equation one degree less than (2), and consequently two degrees less than (1). Let this new equation be represented by

3

3

4

+Cn-3x+Cn-2=0. (3)

If a3 is a root of equation (3), we can divide it by x-aз, we shall thus find a new equation of three degrees less than equation (1). If we continue in this way, we shall, after n divisions, obtain an equation whose degree =0; therefore, equation (1) is composed of n factors

X ·ɑ1; x−ɑ ɔ; x−α3; &c.

Hence, we have the following property :

(241.) If a1, A2, A3,

general equation of the nth degree, then, this equation will take

This equation is verified by making either of the n factors=0; that is, by making x=a1, or x=α2, or x=a3, &c., from which we infer, that every equation of the nth degree has

n roots.

(242.) It does not however follow, that all the roots a1, a2, aз, α4, &c., are different, since two or more of them may be equal, but still their number must be n since there are n factors.

will be positive, consequently each term of its equivalent

If the roots are all positive, then will the terms of

(x-ai)(x-a2)(x-a3) (x-An-1)(x—an)=0, when expanded, be alternately positive and negative.

(244.) Hence, if the terms of any equation are neither all positive, nor alternately positive and negative, that equation must contain both positive and negative roots.

(245.) Reasoning after this manner, Des Cartes has shown,

That every equation whose roots are possible, has as many changes of signs from + to -, or from – to +, as there are positive roots ; and as many continuations of the same signs from + to +, or from – to –, as there are negative roots.

(246.) If, as we have already supposed, the n roots of an equation of the nth degree be denoted by a, a,, ag, ---- One we can put the equation under the following form:

(r-a1)(x-a2)(x-a3)(x-ar) (c-am)=0. (1) Let us suppose aj >az; az/az; az >a4 ; and so of the

rest.

If a quantity b greater than dį be substituted for x in (1), the result will be positive, since all the factors will then be positive.

If a quantity c less than an, but greater than an, be substituted for 2, the factors will be all positive except one, and consequently the result will be negative.

If a quantity d less than az, but greater than az, be substituted for x, all the factors except two will be positive; and since two negative factors produce a positive product, the result must be positive.

By following out this plan of reasoning, we deduce the following property :

02

(247.) If two quantities be successively substituted for x in any equation, and give results affected with DIFFERENT SIGNS, there must be an odd number of roots between these quantities.

But if the two quantities when substituted for x give results affected with the SAME SIGNS, there must be either no root, or else an even number of roots between these quantities.

EXAMPLES.

1. Find the first figure of one of the roots of the equation

Q3 +1.5x2 +0.3x—46=0. If we substitute 3 for x, the result will be -4.6, a negative quantity. If we substitute 4 for X, the result will be 43.2, a positive quantity. Therefore, the first figure of the root sought must be 3.

2. Find the first figure of one of the roots of the equation

x 4 + 3x3 +2x2 +6x-148=0. Putting 2 for X, the result is -88, and putting 3 for x, we get 50, ... the first figure of the root sought is 2. 3. Find the first figure of one of the roots of the equation

X3–17x2 +54x-350=0. In this example, the two consecutive numbers between which there is a root, are 10 and 20, therefore, the first figure of the root sought is 1 in the tens' place. (248.) By actual multiplication, we find

-ai (2-a1)(x-22)=x_

Sætajam, 02 za1

taa2 (c-a)(x-a2)(x-a3)=a, x2 +2103 x0102139

2

}+

)